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## Estimating Distinct Elements, Optimally

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### Estimating Distinct Elements, Optimally

David Woodruff

IBM

Based on papers with Piotr Indyk, Daniel Kane, and Jelani Nelson

Problem Description

- Given a long string of at most n distinct characters, count the number F0 of distinct characters
- See characters one at a time
- One pass over the string
- Algorithms must use small memory and fast update time
- too expensive to store set of distinct characters
- algorithms must be randomized and settle for an approximate solution: output F 2 [(1-²)F0, (1+²)F0] with, say, good constant probability

Algorithm History

- Flajolet and Martin introduced problem, FOCS 1983
- O(log n) space for fixed ε in random oracle model
- Alon, Matias and Szegedy
- O(log n) space/update time for fixed ε with no oracle
- Gibbons and Tirthapura
- O(ε-2 log n) space and O(ε-2) update time
- Bar-Yossef et al
- O(ε-2 log n) space and O(log 1/ε) update time
- O(ε-2log log n + log n) space and O(ε-2) update time, essentially
- Similar space bound also obtained by Flajolet et al in the random oracle model
- Kane, Nelson and W
- O(ε-2+ log n) space and O(1) update and reporting time
- All time complexities are in unit-cost RAM model

Lower Bound History

- Alon, Matias and Szegedy
- Any algorithm requires Ω(log n) bits of space
- Bar-Yossef
- Any algorithm requires Ω(ε-1) bits of space
- Indyk and W
- If ε > 1/n1/9, any algorithm needs Ω(ε-2) bits of space
- W
- If ε > 1/n1/2, any algorithm needs Ω(ε-2) bits of space
- Jayram, Kumar and Sivakumar
- Simpler proof of Ω(ε-2) bound for any ε > 1/m1/2
- Brody and Chakrabarti
- Show above lower bounds hold even for multiple passes over the string

Combining upper and lower bounds, the complexity of this problem is:

Θ(ε-2+ log n) space and Θ(1) update and reporting time

Outline for Remainder of Talk

- Proofs of the Upper Bounds
- Proofs of the Lower Bounds

Hash Functions for Throwing Balls

- We consider a random mapping f of B balls into C containers and count the number of non-empty containers
- The expected number of non-empty containers is C – C(1-1/C)B
- If instead of the mapping f, we use an O(log C/ε)/log log C/ε – wise independent mapping g, then
- the expected number of non-empty containers under g is the same as that under f, up to a factor of (1 ± ε)
- Proof based on approximate inclusion-exclusion
- express 1 – (1-1/C)B in terms of a series of binomial coefficients
- truncate the series at an appropriate place
- use limited independence to handle the remaining terms

Fast Hash Functions

- Use hash functions g that can be evaluated in O(1) time.
- If g is O(log C/ε)/(log log C/ε)-wise independent, the natural family of polynomial hash functions doesn’t work
- We use theorems due to Pagh, Pagh, and Siegel that construct k-wise independent families for large k, and allow O(1) evaluation time
- For example, Siegel shows:
- Let U = [u] and V = [v] with u = vc for a constant c > 1, and suppose the machine word size is Ω(log v)
- Let k = vo(1) be arbitrary
- For any constant d > 0 there is a randomized procedure that constructs a k-wise independent hash family H from U to V that succeeds with probability 1-1/vd and requires vd space. Each h 2 H can be evaluated in O(1) time
- Can show we have sufficiently random hash functions that can be evaluated in O(1) time and represented with O(ε-2+ log n) bits of space

Algorithm Outline

- Set K = 1/ε2
- Instantiate a lg n x K bitmatrix A, initializing entries of A to 0
- Pick random hash functions f: [n]->[n] and g: [n]->[K]
- Obtain a constant factor approximation R to F0 somehow
- Update(i): Set A1, g(i) = 1, A2, g(i) = 1, …, Alsb(f(i)), g(i) = 1
- Estimator: Let T = |{j in [K]: Alog (16R/K), j = 1}|

Output (32R/K) * ln(1-T/K)/ln(1-1/K)

Space Complexity

- Naively, A is a lg n x K bitmatrix, so O(ε-2 log n) space
- Better: for each column j, store the identity of the largest row i(j) for which

Ai, j = 1. Note if Ai,j = 1, then Ai’, j = 1 for all i’ < i

- Takes O(ε-2 log log n) space
- Better yet: maintain a “base level” I. For each column j, store max(i(j) – I, 0)
- Given an O(1)-approximation R to F0 at each point in the stream, set

I = log R

- Don’t need to remember i(j) if i(j) < I, since j won’t be used in estimator
- For the j for which i(j) ¸ I, about 1/2 such j will have i(j) = I, about one fourth such j will have i(j) = I+1, etc.
- Total number of bits to store offsets is now only O(K) = O(ε-2) with good probability at all points in the stream

The Constant Factor Approximation

- Previous algorithms state that at each point in the stream, with probability 1-δ, the output is an O(1)-approximation to F0
- The space of such algorithms is O(log n log 1/δ).
- Union-bounding over a stream of length m gives O(log n log m) total space
- We achieve O(log n) space, and guarantee the O(1)-approximation R of the algorithm is non-decreasing
- Apply the previous scheme on a log n x log n/(log log n) matrix
- For each column, maintain the identity of the deepest row with value 1
- Output 2i, where i is the largest row containing a constant fraction of 1s
- We repeat the procedure O(1) times, and output the median of the estimates
- Can show the output is correct with probability 1- O(1/log n)
- Then we use the non-decreasing property to union-bound over O(log n) events
- We only increase the base level every time R increases by a factor of 2
- Note that the base level never decreases

Running Time

- Blandford and Blelloch
- Definition: a variable length array (VLA) is a data structure implementing an array C1, …, Cn supporting the following operations:
- Update(i, x) sets the value of Ci to x
- Read(i) returnsCi

The Ci are allowed to have bit representations of varying lengths len(Ci).

- Theorem: there is a VLA using O(n + sumi len(Ci)) bits of space supporting worst case O(1) updates and reads, assuming the machine word size is at least log n
- Store our offsets in a VLA, giving O(1) update time for a fixed base level
- Occasionally we need to update the base level and decrement offsets by 1
- Show base level only increases after Θ(ε-2) updates, so can spread this work across these updates, so O(1) worst-case update time
- Copy the data structure, use it for performing this additional work so it doesn’t interfere with reporting the correct answer
- When base level changes, switch to copy
- For O(1) reporting time, maintain a count of non-zero containers in a level

Outline for Remainder of Talk

- Proofs of the Upper Bounds
- Proofs of the Lower Bounds

1-Round Communication Complexity

Alice:

Bob:

What is f(x,y)?

input x

input y

- Alice sends a single, randomized message M(x) to Bob
- Bob outputs g(M(x), y) for a randomized function g
- g(M(x), y) should equal f(x,y) with constant probability
- Communication cost CC(f) is |M(x)|, maximized over x and random bits
- Alice creates a string s(x), runs a randomized algorithm A on s(x), and
- transmits the state of A(s(x)) to Bob
- Bob creates a string s(y), continues A on s(y), thus computing A(s(x)◦s(y))
- If A(s(x)◦s(y)) can be used to solve f(x,y), then space(A) ¸ CC(f)

The Ω(log n) Bound

- Consider equality function: f(x,y) = 1 if and only if x = y for x, y 2 {0,1}n/3
- Well known that CC(f) = Ω(log n)for(n/3)-bit stringsxandy
- Let C: {0,1}n/3 -> {0,1}nbe an error-correcting code with all codewords of Hamming weight n/10
- If x = y, then C(x) = C(y)
- If x != y, then¢(C(x), C(y)) = Ω(n)
- Let s(x) be any string on alphabet size n with i-th character appearing in s(x) if and only if C(x)i = 1. Similarly define s(y)
- If x = y, then F0(s(x)◦s(y)) = n/10. Else, F0(s(x)◦s(y)) = n/10 + Ω(n)
- A constant factor approximation to F0solves f(x,y)

The Ω(ε-2) Bound

- Let r = 1/ε2. Gap Hamming promise problem for x, y in {0,1}r
- f(x,y) = 1 if ¢(x,y) > 1/(2ε-2)
- f(x,y) = 0 if ¢(x,y) < 1/(2ε-2) - 1/ε
- Theorem: CC(f) = Ω(ε-2)
- Can prove this from the Indexing function
- Alice has w 2 {0,1}r, Bob has i in {1, 2, …, r}, output g(w, i) = wi
- Well-known that CC(g) = Ω(r)
- Proof: CC(f) = Ω(r),
- Alice sends the seed r of a pseudorandom generator to Bob, so the parties have common random strings zi, …, zr2 {0,1}r
- Alice sets x = coordinate-wise-majority{zi | wj = 1}
- Bob sets y = zi
- Since the ziare random, if xj = 1, then by properties of majority, with good probability ¢f(x,y) < 1/(2ε-2) - 1/ε, otherwise likely that¢f(x,y) > 1/(2ε-2)
- Repeat a few times to get concentration

The Ω(ε-2) Bound Continued

- Need to create strings s(x) and s(y) to have F0(s(x)◦s(y)) decidewhether¢(x,y) > 1/(2ε-2) or ¢(x,y) < 1/(2ε-2) - 1/ε
- Let s(x) be a string on n characters where character i appears if and only if xi = 1. Similarly define s(y)
- F0(s(x)◦s(y)) = (wt(x) + wt(y) + ¢(x,y))/2
- Alice sends wt(x) to Bob
- A calculation shows a (1+ε)-approximation to F0(s(x)◦s(y)), together with wt(x) and wt(y), solves the Gap-Hamming problem
- Total communication is space(A) + log 1/ε = Ω(ε-2)
- It follows that space(A) = Ω(ε-2)

Conclusion

Combining upper and lower bounds, the streaming complexity of

estimating F0 up to a (1+ε) factor is:

Θ(ε-2+ log n) bits of space and Θ(1) update and reporting time

- Upper bounds based on careful combination of efficient hashing,
- sampling and various data structures
- Lower bounds come from 1-way communication complexity

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